Suppose we have two schemes X,Y and a map f:X→Y. Then we know that HomX(f∗G,F)≃HomY(G,f∗F), where F is an OX-module and G an OY-module (and the Homs are in the category of OX-modules etc). This gives a natural map f∗f∗F→F, just by setting G=f∗F and looking at where the identity map goes.

Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if F is a very ample invertible sheaf (and maybe some more hypothesis on the map and X and Y were assumed as well).

**Answer**

This question may have general interest to students: What can be said about the canonical map

ρ:π∗π∗E→E?

where π:X→S any map of schemes and E a quasi coherent OX-module?

Example. Let S:=Spec(k) with k a field and let X be a scheme of finite type over S.

If E is a quasi coherent OX-module and π:X→S is the canonical map, it follows π∗π∗E→E

is the following map:

M1. ρ:OX⊗H0(X,E)→E.

defined by ρ(U)(e⊗sU):=esU∈E(U) on the open set U. Here e∈OX(U) and sU is the restriction of the global section s to the open set U. Since H0(X,E) is a k-vector space it follows π∗π∗E is a free OX-module of rank dimk(H0(X,E)).

Example.If E is a rank r locally trivial OX-module

we may associate a “geometric vector bundle” π:V(E∗)→X to E, and the fiber π−1(p)≅Arκ(p) is affine r-space over κ(p) – the residue field of p. The fibration π is by Hartshorne, Ex. II.5.18 a Zariski locally trivial fibration with affine spaces as fibers.

The map ρ “evaluates” a global section s in the fiber E(p) at a point p∈X. We get for every p an element ρ(p)(s)∈E(p). Here Arκ(p)≅Spec(Symκ(p)(E(p)∗))

Hence the map ρ is surjective iff E is generated by global sections. I believe the following claim is correct:

Lemma. Let k=A be any commutative ring with Γ(X,E) a free A-module. If the map ρ is an isomorphism it follows E is a free (or trivial) OX-module.

Proof. Let S:=Spec(k) and let π:X→S be the structure morphism. There is by definition a map

M1. π#:OS→π∗OX

and this gives a canonical map

ρ:k:=OS(S)→Γ(X,OX).

Hence Γ(X,OX) is a k-algebra and left k-module.

It follows ρ gives the abelian group Γ(X,E) the structure of a k-module with a basis {ei}i∈I. Hence π∗E is a trivial OS-module on the basis {ei}i∈I. The pull back of a locally trivial OS-module is a locally trivial OX-module, and it follows there is an isomorphism

M2. π∗π∗E≅π∗(⊕i∈IOSei)≅⊕i∈IOXei

since pull back commutes with direct sums and there is an isomorphism

π∗(OS)≅OX⊗π−1(OS)π−1(OS)≅OX

It follows π∗π∗E≅⊕i∈IOXei is a free OX-module on the set {ei}i∈I. Hence if ρ is an isomorphism it follows E is a free OX-module. QED

A comment on “free/trivial” modules. There seems to be confusion on what is meant by “free (or trivial) OX-module”. I use the following definition:

Definition. Given any set T:={ei}i∈I. We say a left OX-module E is “free (or trivial) on the set T” iff for any open set U⊆X it follows

Free1. E(U)≅⊕i∈IOX(U)ei.

The sum is the direct sum.

It has the following restriction maps: Given V⊆U open subsets, it follows the restriction map

Free2. ηU,V:E(U)→E(V)

is defined by ηU,V(∑i∈Isiei):=⊕i∈I(si)Vei.

Here si∈OX(U) and (si)V∈OX(V) is the restriction of the section si to the open set V. One checks E is a quasi coherent OX-module in general.

I believe with this definition of “free/trivial”, the proof of the “Lemma/observation” above is correct. I agree – it is not difficult, but some students are not used to “sheaves” and OX-modules.

PS: An OX-module E is called “locally trivial of rank r” iff there is an open cover {Ui}i∈I with EUi≅(OX)rUi for all i. The module E is “trivial of rank r” iff we may choose I:={1} and U1:=X. Hence when people use the notion “E is a trivial OX-module” this means E is a direct sum of copies of the structure sheaf OX.

Example: Let k be a field and let B:=k[x],E:=⊕i∈Ik[x]ei,S:=Spec(k) and C:=Spec(B)

Let π:S→S be the canonical map. It follows

π∗π∗E≅B⊗k(⊕i∈IBei)≅⊕i∈IB⊗kBei≅⊕i∈Ik[x,y]ei

and the canonical map

ϕ:B⊗kE→E defined by b⊗e:=be is not an isomorphism in this case since clearly k[x,y]≠k[x].

“If F=OX, then the stereotypical example of when this is true is when X=PnY. More generally (but still with the F=OX), the condition that “f is proper and all fibers are geometrically connected” may suffice, although I am not at all confident of that.”

The fiber of the morphism p:PnY→Y over a point s (with residue field κ(s):=k)

is the space Pnk and projective space is “geometrically connected” in general. Let K be the algebraic closure of k. There is an isomorphism Spec(K)×Spec(k)Pnk≅PnK, and PnK is connected for any field k.

If E is a finite rank locally trivial OX-module

we may consider π:X:=P(E∗)→Y. If Y:=Spec(k) with k and F is a locally trivial OX-module that is not free it follows by the above Lemma that the map ρ:Γ(X,F)⊗OX→E is not an isomorphism.

The answer is: The canonical map ρ is by the above argument(s) seldom an isomorphism.

**Attribution***Source : Link , Question Author : Matt , Answer Author : hm2020*